H₂SO₄:
V=0,95L
Cm=0,420mol/L
n = CmV = 0,42mol/L * 0,95L = 0,399mol
KOH:
V=0,9L
Cm=0,26mol/L
n = CmV = 0,26mol/L * 0,9L = 0,234mol
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
1mol : 2mol
0,399mol : 0,234mol
limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol
n = 0,282mol
V = 0,950L + 0,900L = 1,85L
Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
Hydrogen is usually –1. This is INCORRECT. The oxidation number for H is +1.
Oxygen is usually –2. This is CORRECT.
A pure group 1 element is +1. This is INCORRECT. It does not follow. This will depend on the other elements and the overall charge.
A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Saturated Solution: A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom. Unsaturated Solution: A solution ( with less solute than the saturated solution )that completely dissolves, leaving no remaining substances. Supersaturated Solution.
Answer:
0.179kg/m3 hope that helps you out
